Lösungsvorschlag[Bearbeiten | Quelltext bearbeiten]

Mantelfläche eines Rotationskörpers (siehe Buch S.257-258):

$M=2\pi \int _{a}^{b}f(x){\sqrt {1+f'(x)^{2}}}dx$

Funktion ist linear mit Anstieg ${\frac {r}{h}}$

$f(x)={\frac {r}{h}}x,\quad f'(x)={\frac {r}{h}}$

$M=2\pi \int _{0}^{h}{\frac {r}{h}}x{\sqrt {1+\left({\frac {r}{h}}\right)^{2}}}\,dx$

$=2\pi {\frac {r}{h}}{\sqrt {1+\left({\frac {r}{h}}\right)^{2}}}\int _{0}^{h}x\,dx$

$=2\pi {\frac {r}{h}}{\sqrt {1+\left({\frac {r}{h}}\right)^{2}}}h^{2}{\frac {1}{2}}$

$=\pi {\frac {r}{h}}{\sqrt {1+\left({\frac {r}{h}}\right)^{2}}}h^{2}$

$=\pi {\frac {r}{h}}{\sqrt {{\frac {h^{2}}{h^{2}}}+{\frac {r^{2}}{h^{2}}}}}h^{2}$

$=\pi {\frac {r}{h}}{\frac {\sqrt {r^{2}+h^{2}}}{h}}h^{2}$

$=\pi r{\sqrt {r^{2}+h^{2}}}$

$=\pi rs\quad$ (weil $s={\sqrt {r^{2}+h^{2}}}$ )

Verweise[Bearbeiten | Quelltext bearbeiten]